Let me first recall that the classical Radon transform takes a (smooth compactly supported, say) function $f$ defined on $\mathbb{R}^n$ as an input, and gives as output the map $H\mapsto \int_H f$ for $H$ running over the set of affine hyperplanes. Radon's inversion theorem gives a formula to recover $f$ from its transform.

I needed a similar result in a slightly different context in one of my project. We are on a simply connected, non-positively curved Riemannian manifold $X$ (or more generally a Hadamard space) and we have compactly supported smooth (positive if needed) functions $f$ and $g$. For all geodesic $\gamma\subset X$ (which is globally minimizing thanks to the curvature assumption, in particular it is convex), the metric projection $p_\gamma:X\to\gamma$ is well defined and $1$-Lipschitz. Let us call a perpendicular to $\gamma$ any level of $p_\gamma$. If we know that $\int_P f = \int_P g$ for all perpendicular $P$ of all geodesics, can we deduce that $f=g$?

In fact, I managed to get a result along these lines which is sufficient for my needs; my question is: is this well-known? Does it have a name? Do you know a reference?

Note that in the case of $\mathbb{R}^n$ or of the real hyperbolic space, then perpendiculars are totally geodesic hyperplanes so that we are reduced to the usual Radon transform, and the book by Helgason contains more than I really need. But even when $X$ is the complex hyperbolic space, then the perpendiculars are not the subsets used in the usual Radon transform on symmetric spaces (note that they are not convex).

Note also that the case of trees is an easy but nice combinatorial exercise; I could not find a reference either (I only found a paper studying the Radon transform defined using horospheres).

  • $\begingroup$ I added the classical analysis tag since the arXiv ca includes harmonic as a subset. $\endgroup$ – Willie Wong Jul 14 '10 at 10:20
  • $\begingroup$ Also, do you mind sending me a copy when you have a pre-print ready for this result? It is really interesting. $\endgroup$ – Willie Wong Jul 14 '10 at 10:23
  • $\begingroup$ @Willie: I sure don't mind, but you may be disappointed: I in fact managed to need only the case of finitely supported (hence discontinuous) functions, which is rather unsatisfactory in most ways. $\endgroup$ – Benoît Kloeckner Jul 14 '10 at 11:51
  • $\begingroup$ Dear Benoit, Just a question : what's the dimension of the "manifold" of perpendiculars in a general Hadamard space? In the standard Radon or Horospherical transform you pass from functions defined on an $n$-dimensional manifold to functions defined on a manifold of the same dimension. There is just enough room to get injectivity. I think you may have a lot more room to play with. $\endgroup$ – alvarezpaiva Mar 1 '12 at 10:08
  • $\begingroup$ @alvarezpaiva In the case of a manifold, the dimension is the same (recall that in the case of a Euclidean space, this is just the usual Radon transform). So in general, we do not have much more room. $\endgroup$ – Benoît Kloeckner Mar 2 '12 at 10:29

If we work with Hadamard spaces, a perpendicular to a geodesic needn't be a geodesic, so I wonder what measure you use in the integral.

  • $\begingroup$ This should be a comment, but you're right. The point is that, in fact, the real transform I need to consider is $R\mu: \gamma\mapsto (p_\gamma)_\#(\mu)$ where $\mu$ is a probability measure on $X$ and $\#$ denotes the push-forward. In the case of Hadamard manifold, it is therefore to be understood that the perpendiculars to $\gamma$ are endowed with the measures obtained by disintegrating the volume measure with respect to $p_\gamma$. In the case of Hadamard spaces, such an interpretation is usually not available and one should stick to measures. $\endgroup$ – Benoît Kloeckner Jul 15 '10 at 7:51
  • $\begingroup$ It should really have been a comment, sorry. $\endgroup$ – m b Jul 16 '10 at 7:33
  • $\begingroup$ @m b: It's not you're fault. With the amount of experience points you have, it's not possible to make a comment. You will be able to create comments in the near future, if you spend more time on MO, asking useful questions/and or answering ones. $\endgroup$ – Max Muller Jan 27 '11 at 13:10

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